Question

import urllib2    

def GetBrowserHtml_content(url):
    req_header = {'User-Agent':'Mozilla/5.0 (Windows NT 6.1) AppleWebKit/537.11 (KHTML, like Gecko) Chrome/23.0.1271.64 Safari/537.11',
                  'Accept':'text/html;q=0.9,*/*;q=0.8',
                  'Accept-Charset':'ISO-8859-1,utf-8,gbk;q=0.7,*;q=0.3',
                  'Connection':'close',
                  'Referer':None
                  }
    req_timeout = 5
    request = urllib2.Request(url,None,req_header)
    response = urllib2.urlopen(request,None,req_timeout)
    html_content = response.read()
    return html_content
url = 'http://www.ccdi.gov.cn/jlsc/index_4.html'
html_content = GetBrowserHtml_content(url)

我有一段代码如上所述。

当我运行代码时，我收到以下错误。

Traceback (most recent call last):


File "E:/Programming/python/CWSeg/spider/hahahha.py", line 34, in <module>
    html_content = GetBrowserHtml_content(url)
  File "E:/Programming/python/CWSeg/spider/hahahha.py", line 22, in GetBrowserHtml_content
    response = urllib2.urlopen(request,None,req_timeout)
  File "C:\Python27\lib\urllib2.py", line 126, in urlopen
    return _opener.open(url, data, timeout)
  File "C:\Python27\lib\urllib2.py", line 406, in open
    response = meth(req, response)
  File "C:\Python27\lib\urllib2.py", line 519, in http_response
    'http', request, response, code, msg, hdrs)
  File "C:\Python27\lib\urllib2.py", line 444, in error
    return self._call_chain(*args)
  File "C:\Python27\lib\urllib2.py", line 378, in _call_chain
    result = func(*args)
  File "C:\Python27\lib\urllib2.py", line 527, in http_error_default
    raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 521:

谁能指出我做错了什么？提前谢谢。

urllib2.HTTPError：HTTP错误521：

0 个答案: